At 650 K, the reaction MgCO3(s) ⇌ MgO(s) + CO2(g) has Kp = 0.026....

Question

At 650 K, the reaction MgCO3(s) ⇌ MgO(s) + CO2(g) has Kp = 0.026. A 10.0-L container at 650 K has 1.0 g of MgO(s) and CO2 at P = 0.0260 atm. The container is then compressed to a volume of 0.100 L. Find the mass of MgCO3 that is formed.

Accepted Answer

Hi everyone for this problem, we're told to consider the following reaction. At 1073 kelvin, a reaction vessel with a volume of 4.55 liters contains 2.5 g of calcium oxide and carbon dioxide with a pressure of 1.168 cm, calculate the mass of calcium carbonate formed when the vessel is compressed to a final volume of 0.55 liters. Okay, so our goal here is to calculate the mass of calcium carbonate. Let's take a look at what we're given. We are given an initial volume, an initial pressure and a final volume. So what we can do now is we can use Boyle's Law to find that pressure of our second our our final reaction to create our ice table. So let's go ahead and do that. So we're going to use Boyle's Law, which is P one times V one equals P two times V two. And we don't know what P two is. So let's go ahead and rearrange this to solve for P two. So we get P two is equal to P one times V one over B two. So we can plug in what we know so that we can get P. Two. We know that P one is 1.16 A T. M. And V one is 4.55 liters And we know that our V2 which is our final volume Is 0.455 L. Okay, so once we do that we get 11.6 a. T. M. This is going to be our initial pressure and let's go ahead and write out our reaction rewrite it out so we can create an ice table. So we have our calcium carbonate. Is that equilibrium? And then we have calcium oxide plus carbon dioxide gas. So we're going to create an ice table and I'm going to draw this line here to separate our products from our reactant. So this initial pressure of 11.6 that we just solved for, we're going to assume that it is our carbon dioxide. Okay, So we'll go ahead and ignore everything else here and our change in concentration is going to be minus X. And then we have 11.6 minus X. Ok, so there's 11.6 minus X is going to represent our equilibrium pressure for our carbon dioxide. Our goal here is to solve for X so that we know what X is and we can solve for X by setting R K p equal to the equilibrium pressure of our carbon dioxide. And the problem, they tell us what our K P is. It's right here, they tell us A K p is equal to 1. A. T. M. And our equilibrium pressure of carbon dioxide is 11.6 minus X. That comes from our equilibrium role of our ice table. So, when we saw for X, we get X is equal to 10. A T. M. Okay, so now that we know what our X value is, we can use our ideal gas law to solve for moles of X. Ok, because right now we're in A T. M. And we want moles. So our ideal gas law is PV equals and R. T. And since we want to solve for moles of carbon dioxide, we're going to rearrange this to solve for moles. So we get moles is equal to P V over R. T. So we can plug in what we know. We just solved for p. r pressure and we said that it is 10.44 A. T. M. R volume Is going to be 4.55 L. Our is our gas constant. And this is a value. We should have memorized. This is 0. leaders times atmosphere over more times kelvin times our temperature of 1073 kelvin this was given. So we're going to make sure our units cancel properly here and we should be left with just moves. So our A. T. M cancel, our leaders cancel and our kelvin cancels. So we're left with 0.5395 moles of carbon dioxide. Okay, so now we need to figure out how many moles of calcium oxide we have because we have to react ints and we need to figure out which one is the limiting reactant. So we know our moles of CO2. So let's calculate our moles of calcium oxide. So in the problem we're told we have 2.5 g of calcium oxide. So we need to go from moles of calcium oxide or we need to go from grams of calcium oxide, two moles of calcium oxide. And we can do that using molar mass. So one mole of calcium oxide Is 56.078 g of calcium oxide. Our grams of calcium oxide cancel. And we have zero . moles of calcium oxide. So let's compare. The two are moles of carbon dioxide and moles of calcium oxide are moles of calcium oxide is less. So that means that this is going to be our limiting reactant. It is going to run out first. So now that we know are molds of limiting reactant. We can use our multiple ratio between calcium oxide and calcium carbonate to find our molar mass or our mass of calcium carbonate, Which is what the question is asking for. So let's go ahead and rewrite our molds for our limiting reactant. So we have zero point 04458 moles of calcium oxide. Using our multiple ratio and looking at our reaction Up at top, we see we have a 1-1 molar ratio between calcium carbonate and calcium oxide. So let's go ahead and write that in. So for every one mole of calcium carbonate, we have one Mall of Calcium Oxide. So our moles of calcium oxide cancel. And we have moles of calcium carbonate. The question is asking us for the mass of calcium carbonate. So we can go from moles of calcium carbonate to mass of calcium carbonate using molar mass. So in one mole of calcium carbonate, using our periodic table, That is .088g of calcium carbonate. Okay, so our molds of calcium carbonate cancel. And we're left with grams of calcium carbonate. So we're going to get a final answer of 4.46 g of calcium carbonate as our final answer. Okay, and that is the end of this problem. I hope this was helpful.

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